Posted by briansmccabe on 12/19/05 21:50

my code:

$query6 = "SELECT COUNT (movie_id) as quant FROM movies WHERE divx = 1
AND format = 'reg'";
$result6 = mysql_fetch_array($query6);

later on the page:

echo $result6[quant]

result:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
result resource in /home/tgupc/public_html/admin/divxreport.php on line
19


Any suggestions?

thanks

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